Existence and simulation of positive solutions for m-point fractional differential equations with derivative terms

Wenchao Sun; You-Hui Su; Ai Sun; Quanxing Zhu

doi:10.1515/math-2021-0131

Article Open Access

Existence and simulation of positive solutions for m-point fractional differential equations with derivative terms

Wenchao Sun , You-Hui Su , Ai Sun and Quanxing Zhu

Published/Copyright: December 31, 2021

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$Open Mathematics$

From the journal Open Mathematics Volume 19 Issue 1

Abstract

In this article, we investigate the existence of positive solutions for a class of m -point fractional differential equations whose nonlinear terms involve derivatives. By using the properties of the Green function and fixed point theorems, some new conditions for the existence of at least one, two, three, and 2 n − 1 positive solutions are established. As verification, some simulation examples are given to illustrate the main results. It is worth mentioning that we study the m -point fractional differential equations with nonlinear terms involving derivative and use the iterative method to simulate our examples and give the approximate solution.

Keywords: fractional differential equations; existence; positive solution; simulation

MSC 2010: 34A08; 34B18; 34K37

1 Introduction

In recent years, the boundary value problem of differential equations have become a research hotspot in physics, chemistry, engineering, and other fields, see [1,2,3, 4,5]. However, with the in-depth study of nonlinear differential equations, people found that integer order differential equations have many limitations in practical applications, such as the description of capacitance and inductance in physics, the melting of polymer materials in chemistry, and so on, and the description of fractional differential equations is more realistic than the ideal model of integer order. This has attracted more and more scholars to study fractional differential equations and have obtained some important results, see [6,7, 8,9,10, 11,12] and references therein. In [13], Jia et al. investigated the positive solutions to the following boundary value problems of fractional differential equations

D 0 + α ( p ( t ) D 0 + β u ( t ) ) + λ f ( t , u ( t ) ) = 0 , 0 < t < 1 , lim t → 0 + t 2 − β u ( t ) = a , u ( 1 ) = ∫ 0 1 u ( s ) d A ( s ) , lim t → 0 + t 1 − α p ( t ) D 0 + β u ( t ) = 0 ,

where 0 < α ≤ 1 , 1 < β ≤ 2 , λ > 0 , a ≥ 0 , and f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , D 0 + α and D 0 + β are the standard Riemann-Liouville fractional derivatives. By using fixed point index theory, the existence of at least one, two and the nonexistence of positive solutions are obtained. However, the author only considered the case that the nonlinear terms have no derivative. As we all know, the nonlinear terms with derivative of the differential equations have more practical significance, see [14,15, 16,17]. In [17], Sheng et al. studied the following boundary value problems of fractional differential equations

D 0 + α u ( t ) + f ( t , u ( t ) , D 0 + β u ( t ) ) = 0 , 0 < t < 1 , u ( 0 ) = u ′ ( 0 ) = u ′ ( 1 ) = 0 ,

where 0 < β < 1 , 2 < α < 3 , D 0 + α and D 0 + β are the standard Riemann-Liouville fractional derivatives. By using the Avery-Peterson fixed point theorem, the results that there are at least three solutions are obtained.

In the literature mentioned above, the authors merely researched the existence of solutions in theoretical aspect, yet, its numerical simulation part is rarely involved, see [18,19, 20,21] and references therein. In [21], Ahmad et al. concerned the following numerical treatment of three-term time fractional-order multi-dimensional diffusion equations

a 1 ∂ α U ( y ¯ , t ) ∂ t α + a 2 ∂ β U ( y ¯ , t ) ∂ t β + a 3 ∂ γ U ( y ¯ , t ) ∂ t γ = a 4 ∇ U ( y ¯ , t ) + F ( y ¯ , t ) ≡ £ U ( y ¯ , t ) , U ( y ¯ , 0 ) = U 0 , C ( y ¯ , t ) = g ( y ¯ , t ) , y ¯ ∈ ∂ Ω ,

where 0 < γ ≤ β ≤ α ≤ 1 , t > 0 , a 4 is the diffusion coefficient and for one-dimensional (1D) case y ¯ = x , for two-dimensional (2D) case y ¯ = ( x , y ) , and for three-dimensional (3D) case y ¯ = ( x , y , z ) . By using an efficient local meshless method, the numerical results are obtained for 1D, 2D, and 3D cases on rectangular and nonrectangular computational domains. Inspired by the above literature, we study the following m -point fractional differential equations whose nonlinear terms involve derivative:

(1.1) D 0 + α u ( t ) + λ f ( t , u ( t ) , D 0 + β u ( t ) ) = 0 , t ∈ ( 0 , 1 ) , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 2 ) ( 0 ) = 0 , D 0 + β u ( 1 ) = ∑ i = 1 m − 2 ε i D 0 + β u ( η i ) ,

where f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , λ > 0 , n − 1 < α < n , β > 1 , and α − β ≥ 1 , ε i > 0 , 0 < η i < 1 , i = 1 , 2 , … , m − 2 , ∑ i = 1 m − 2 ε i η i α − β − 1 < 1 . D 0 + α , and D 0 + β are the standard Riemann-Liouville fractional derivatives. The existence of at least one, two, three, and 2 n − 1 positive solutions are established by using some fixed point theorems in Banach space. In particular, some simulation examples are given to illustrate the main results.

The paper is organized as follows. In Section 2, we introduce some basic definitions and lemmas and give the properties of Green’s function. In Section 3, the completely continuous operator for fractional differential equation (1.1) is defined. In Section 4, we prove the existence of at least one positive solution for boundary value problems (1.1), and give some simulation examples to illustrate the main results. In Section 5, the existence of two, three, and 2 n − 1 positive solutions for boundary value problems (1.1) are investigated, and some examples are simulated by using the iterative method. In Section 6, conclusions and prospects are given.

2 Preliminaries

In this section, some basic definitions and lemmas are introduced to help us understand the main results and proofs in Sections 3, 4, and 5.

Definition 2.1

[22] The Riemann-Liouville fractional integral of order α > 0 of a function y : ( 0 , + ∞ ) → R is given by

I 0 + α y ( t ) = 1 Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s ,

provided that the right side is pointwise defined on ( 0 , + ∞ ) .

Definition 2.2

[22] The Riemann-Liouville fractional derivative of order α > 0 of a function y : ( 0 , + ∞ ) → R is given by

D 0 + α y ( t ) = 1 Γ ( n − α ) d d t n ∫ 0 t y ( s ) ( t − s ) α − n + 1 d s ,

where, n = [ α ] + 1 , [ α ] denotes the integer part of number α , provided the right side is pointwise defined on ( 0 , + ∞ ) .

Lemma 2.3

[23] The equality D 0 + β I 0 + α u ( t ) = I α − β u ( t ) , α > 0 , β > 0 holds for u ∈ C ( 0 , 1 ) ∩ L ( 0 , 1 ) .

Lemma 2.4

[23] Assume that y ∈ C ( 0 , 1 ) ∩ L ( 0 , 1 ) with a fractional derivative of α > 0 that belongs to C ( 0 , 1 ) ∩ L ( 0 , 1 ) . Then the fractional differential equation

D 0 + α y ( t ) = 0

has a unique solution y ( t ) = C 1 t α − 1 + C 2 t α − 2 + ⋯ + C n t α − n , where C i ∈ R , i = 1 , 2 , … , n , n is the smallest integer greater than or equal to α .

Lemma 2.5

[23] Assume that y ∈ C ( 0 , 1 ) ∩ L ( 0 , 1 ) with a fractional derivative of α > 0 that belongs to C ( 0 , 1 ) ∩ L ( 0 , 1 ) . Then

I 0 + α D 0 + α y ( t ) = y ( t ) + C 1 t α − 1 + C 2 t α − 2 + ⋯ + C n t α − n ,

for some C i ∈ R , i = 1 , 2 , … , n , where n is the smallest integer greater than or equal to α .

Lemma 2.6

Assume y ∈ C [ 0 , 1 ] , ∑ i = 1 m − 2 ε i η i α − β − 1 < 1 , α − β ≥ 1 , the problem

(2.1) D 0 + α u ( t ) + λ y ( t ) = 0 , t ∈ [ 0 , 1 ] , n − 1 < α < n , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 2 ) ( 0 ) = 0 , D 0 + β u ( 1 ) = ∑ i = 1 m − 2 ε i D 0 + β u ( η i ) , ε i > 0 , 0 < η i < 1 , β > 1 ,

has a unique solution

u ( t ) = λ ∫ 0 1 G ( t , s ) y ( s ) d s ,

where

(2.2) G ( t , s ) = g ( s ) t α − 1 ( 1 − s ) α − β − 1 − g ( 0 ) ( t − s ) α − 1 g ( 0 ) Γ ( α ) , 0 ≤ s ≤ t ≤ 1 , g ( s ) t α − 1 ( 1 − s ) α − β − 1 g ( 0 ) Γ ( α ) , 0 ≤ t ≤ s ≤ 1 ,

is Green’s function of problem (2.1), here

g ( s ) = 1 − ∑ i = 1 , η i ≥ s m − 2 ε i η i − s 1 − s α − β − 1 .

Furthermore, we can obtain

D 0 + β u ( t ) = λ ∫ 0 1 H ( t , s ) y ( s ) d s ,

where

(2.3) H ( t , s ) = g ( s ) t α − β − 1 ( 1 − s ) α − β − 1 − g ( 0 ) ( t − s ) α − β − 1 g ( 0 ) Γ ( α − β ) , 0 ≤ s ≤ t ≤ 1 , g ( s ) t α − β − 1 ( 1 − s ) α − β − 1 g ( 0 ) Γ ( α − β ) , 0 ≤ t ≤ s ≤ 1 .

Proof

According to Lemma 2.5, we can obtain

u ( t ) = − λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s + C 1 t α − 1 + C 2 t α − 2 + ⋯ + C n t α − n .

Since

u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 2 ) ( 0 ) = 0 ,

it implies

c 2 = c 3 = ⋯ = c n − 2 = 0 .

Then

(2.4) u ( t ) = − λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s + C 1 t α − 1 .

Therefore,

(2.5) D 0 + β u ( t ) = − λ Γ ( α − β ) ∫ 0 t ( t − s ) α − β − 1 y ( s ) d s + C 1 Γ ( α ) Γ ( α − β ) t α − β − 1 .

By the boundary condition D 0 + β u ( 1 ) = ∑ i = 1 m − 2 ε i D 0 + β u ( η i ) , we have

− λ Γ ( α − β ) ∫ 0 1 ( 1 − s ) α − β − 1 y ( s ) d s + C 1 Γ ( α ) Γ ( α − β ) = ∑ i = 1 m − 2 ε i − λ Γ ( α − β ) ∫ 0 η i ( η i − s ) α − β − 1 y ( s ) d s + C 1 Γ ( α ) Γ ( α − β ) η i α − β − 1 ,

it can be written as

C 1 = λ g ( 0 ) Γ ( α ) ∫ 0 1 ( 1 − s ) α − β − 1 y ( s ) d s − ∑ i = 1 m − 2 ε i ∫ 0 η i ( η i − s ) α − β − 1 y ( s ) d s = λ g ( 0 ) Γ ( α ) ∫ 0 1 ( 1 − s ) α − β − 1 g ( s ) y ( s ) d s .

Substituting C 1 into (2.4) and (2.5), we obtain

u ( t ) = − λ Γ ( α ) ∫ 0 t ( t − s ) α − 1 y ( s ) d s + λ t α − 1 g ( 0 ) Γ ( α ) ∫ 0 1 ( 1 − s ) α − β − 1 g ( s ) y ( s ) d s

= λ ∫ 0 t g ( s ) t α − 1 ( 1 − s ) α − β − 1 − g ( 0 ) ( t − s ) α − 1 g ( 0 ) Γ ( α ) y ( s ) d s + λ ∫ t 1 g ( s ) t α − 1 ( 1 − s ) α − β − 1 g ( 0 ) Γ ( α ) y ( s ) d s = λ ∫ 0 1 G ( t , s ) y ( s ) d s

and

D 0 + β u ( t ) = − λ Γ ( α − β ) ∫ 0 t ( t − s ) α − β − 1 y ( s ) d s + λ t α − β − 1 g ( 0 ) Γ ( α − β ) ∫ 0 1 ( 1 − s ) α − β − 1 g ( s ) y ( s ) d s = λ ∫ 0 t g ( s ) t α − β − 1 ( 1 − s ) α − β − 1 − g ( 0 ) ( t − s ) α − β − 1 g ( 0 ) Γ ( α − β ) y ( s ) d s + λ ∫ t 1 g ( s ) t α − β − 1 ( 1 − s ) α − β − 1 g ( 0 ) Γ ( α − β ) y ( s ) d s = λ ∫ 0 1 H ( t , s ) y ( s ) d s .

The proof is complete.□

Lemma 2.7

Assume ∑ i = 1 , η i ≥ s m − 2 ε i η i α − β − 1 < 1 , the function g ( s ) defined in Lemma 2.6 is nonnegative and monotone nondecreasing on [ 0 , 1 ] .

Proof

According to the function g ( s ) , we can easily obtain

g ′ ( s ) = − ∑ i = 1 , η i ≥ s m − 2 ( α − β − 1 ) ε i η i − s 1 − s α − β − 2 η i − 1 ( 1 − s ) 2 ≥ 0 .

Then, g ( s ) is monotone nondecreasing on [ 0 , 1 ] .

Since ∑ i = 1 , η i ≥ s m − 2 ε i η i α − β − 1 < 1 , we have

g ( s ) ≥ g ( 0 ) = 1 − ∑ i = 1 , η i ≥ s m − 2 ε i η i α − β − 1 > 0 .

Hence, the function g ( s ) is nonnegative and monotone nondecreasing on [ 0 , 1 ] .

The proof is complete.□

Lemma 2.8

The functions G ( t , s ) and H ( t , s ) defined in Lemma 2.6 are continuous on [ 0 , 1 ] × [ 0 , 1 ] , and satisfy the following conditions:

∀ t , s ∈ [ 0 , 1 ] , 0 ≤ t α − 1 G ( 1 , s ) ≤ G ( t , s ) ≤ G ( 1 , s ) ;
∀ t , s ∈ [ 0 , 1 ] , 0 ≤ t α − β − 1 H ( 1 , s ) ≤ H ( t , s ) ≤ H ( 1 , s ) .

Proof

For 0 ≤ t ≤ s ≤ 1 , it is easy to obtain that G ( t , s ) is increasing on t ∈ [ 0 , s ] and G ( 0 , s ) = 0 .

For 0 ≤ s ≤ t ≤ 1 , we have

∂ G ( t , s ) ∂ t = g ( s ) ( α − 1 ) t α − 2 ( 1 − s ) α − β − 1 − g ( 0 ) ( α − 1 ) ( t − s ) α − 2 g ( 0 ) Γ ( α ) ≥ ( α − 1 ) [ g ( s ) t α − 2 ( 1 − s ) α − β − 1 − g ( 0 ) ( t − t s ) α − 2 ] g ( 0 ) Γ ( α ) = ( α − 1 ) t α − 2 [ g ( s ) ( 1 − s ) α − β − 1 − g ( 0 ) ( 1 − s ) α − 2 ] g ( 0 ) Γ ( α ) ≥ 0 .

Then, G ( t , s ) is increasing on t ∈ [ s , 1 ] .

Consequently, G ( t , s ) is increasing on t ∈ [ 0 , 1 ] , which implies that 0 ≤ G ( t , s ) ≤ G ( 1 , s ) .

On the other hand, for 0 ≤ s ≤ t ≤ 1 , we have

G ( t , s ) = g ( s ) t α − 1 ( 1 − s ) α − β − 1 − g ( 0 ) ( t − s ) α − 1 g ( 0 ) Γ ( α ) ≥ t α − 1 [ g ( s ) ( 1 − s ) α − β − 1 − g ( 0 ) ( 1 − s ) α − 1 ] g ( 0 ) Γ ( α ) = t α − 1 G ( 1 , s ) .

When 0 ≤ t ≤ s ≤ 1 , we obtain

G ( t , s ) = g ( s ) t α − 1 ( 1 − s ) α − β − 1 g ( 0 ) Γ ( α ) ≥ t α − 1 G ( 1 , s ) ,

therefore, 0 ≤ t α − 1 G ( 1 , s ) ≤ G ( t , s ) ≤ G ( 1 , s ) for t , s ∈ [ 0 , 1 ] .

In a similar way, we can obtain

0 ≤ t α − β − 1 H ( 1 , s ) ≤ H ( t , s ) ≤ H ( 1 , s ) ,

for t , s ∈ [ 0 , 1 ] .

The proof is complete.□

Lemma 2.9

[24] Let P be a cone in a Banach space E . Assume Ω 1 and Ω 2 are open subsets of E with 0 ∈ Ω 1 and Ω ¯ 1 ⊂ Ω 2 . If A : P ∩ ( Ω ¯ 2 ⧹ Ω 1 ) → P is a completely continuous operator such that either

‖ A x ‖ ≤ ‖ x ‖ , ∀ x ∈ P ∩ ∂ Ω 1 , and ‖ A x ‖ ≥ ‖ x ‖ , ∀ x ∈ P ∩ ∂ Ω 2 , or
‖ A x ‖ ≥ ‖ x ‖ , ∀ x ∈ P ∩ ∂ Ω 1 , and ‖ A x ‖ ≤ ‖ x ‖ , ∀ x ∈ P ∩ ∂ Ω 2 .

Then A has a fixed point in P ∩ ( Ω ¯ 2 ⧹ Ω 1 ) .

Assume that P c = { x ∈ P : ‖ u ‖ ≤ c } , P ( q , b , d ) = { x ∈ P ∣ b ≤ q ( u ) , ‖ u ‖ ≤ d } , K ( γ , c ) = { x ∈ K : γ ( x ) < c } . We list the following fixed-point theorems.

Lemma 2.10

[25] Let K be a cone in Banach space X . α , γ : K → R + be increasing, nonnegative continuous functional, θ : K → R + be a nonnegative continuous functional, that there exist M , c > 0 such that:

θ ( 0 ) = 0 , and γ ( x ) ≤ θ ( x ) ≤ α ( x ) , ‖ x ‖ ≤ M γ ( x ) , for all x ∈ K ¯ ( γ , c ) ;
There exist 0 < a < b < c , for any λ ∈ [ 0 , 1 ] , x ∈ ∂ K ( θ , b ) such that θ ( λ x ) ≤ λ θ ( x ) ;
T : K ¯ ( γ , c ) → K is a completely continuous operator;
γ ( T x ) > c for x ∈ ∂ K ( γ , c ) ; θ ( T x ) < b for x ∈ ∂ K ( θ , b ) ; K ( α , a ) ≠ ∅ , and α ( T x ) > a for x ∈ ∂ K ( α , a ) .

Then T has at least two fixed points x 1 , x 2 belonging to K ( γ , c ) such that

a < α ( x 1 ) , θ ( x 1 ) < b < θ ( x 2 ) , γ ( x 2 ) < c .

Lemma 2.11

[26] Let P be a cone in real Banach space E . Let α , β and γ be increasing, nonnegative continuous functional on P such that for some c > 0 and H > 0 , γ ( x ) ≤ β ( x ) ≤ α ( x ) , and ‖ x ‖ ≤ H γ ( x ) for all x ∈ P ¯ ( γ , c ) . Suppose that there exist positive numbers a , b , and c with a < b < c , and A : P ¯ ( γ , c ) → P is a completely continuous operator such that:

γ ( A x ) < c for x ∈ ∂ P ( γ , c ) ;
β ( A x ) > b for x ∈ ∂ P ( θ , b ) ;
P ( α , a ) ≠ ∅ , and α ( A x ) < a for x ∈ ∂ P ( α , a ) .

Then A has at least three fixed points x 1 , x 2 , and x 3 belonging to P ( γ , c ) such that

0 < α ( x 1 ) < a < α ( x 2 ) , β ( x 2 ) < b < β ( x 3 ) , γ ( x 3 ) < c .

Lemma 2.12

[27] Suppose that A : P ¯ c → P ¯ c is completely continuous and there exists a concave positive functional q on P such that q ( u ) ≤ ‖ u ‖ for u ∈ P ¯ c . Suppose that there exist constants 0 < a < b < d ≤ c such that:

{ u ∈ P ( q , b , d ) : q ( u ) > b } ≠ ∅ and q ( Au ) > b if u ∈ P ( q , b , d ) ;
‖ Au ‖ < a if u ∈ P a ;
q ( Au ) > b for u ∈ P ( q , b , c ) with ‖ Au ‖ > d .

Then, A has at least three fixed points u 1 , u 2 , u 3 such that

‖ u 1 ‖ < a , b < q ( u 2 ) and u 3 > a with q ( u 3 ) < b .

3 Completely continuous operator

In this section, we define a cone and set a completely continuous operator on the cone.

Let

E = { u ∣ u ∈ C [ 0 , 1 ] , D 0 + β u ∈ C [ 0 , 1 ] }

be the Banach space with the norm ‖ u ‖ = ‖ u ‖ 1 + ‖ u ‖ 2 , where

‖ u ‖ 1 = max t ∈ [ 0 , 1 ] ∣ u ( t ) ∣ , ‖ u ‖ 2 = max t ∈ [ 0 , 1 ] ∣ D 0 + β u ( t ) ∣ .

The cone P ∈ E is defined by

P = u ( t ) ∈ E u ( t ) ≥ 0 , u ( t ) + D 0 + β u ( t ) ≥ 1 4 α − 1 ‖ u ‖ , t ∈ 1 4 , 1 .

We define the operator A : P → P as

( Au ) ( t ) = λ ∫ 0 1 G ( t , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s .

According to Lemma 2.6, we obtain that

D 0 + β ( Au ) ( t ) = λ ∫ 0 1 H ( t , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s .

It should be noted that u ( t ) is a solution of boundary value problem (1.1) if and only if u ( t ) is a fixed point of operator A in P .

Lemma 3.1

The operator A : P → P is completely continuous.

Proof

According to Lemma 2.8, for any u ∈ P , we have

( Au ) ( t ) + D 0 + β ( Au ) ( t ) = λ ∫ 0 1 G ( t , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H ( t , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ t α − 1 ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ t α − β − 1 ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ t α − 1 [ max t ∈ [ 0 , 1 ] ∣ ( Au ) ( t ) ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β ( Au ) ( t ) ∣ ] = λ t α − 1 ‖ Au ‖ .

Then, for t ∈ 1 4 , 1 , we can obtain

( Au ) ( t ) + D 0 + β ( Au ) ( t ) ≥ 1 4 α − 1 ‖ Au ‖ .

Hence, Au ∈ P , which implies that A : P → P .

A : P → P is uniformly bounded.

Let

Ω = { u ∈ P : ‖ u ‖ ≤ R } .

Because f is continuous, for any u ∈ Ω , ( t , u ( t ) , D 0 + α u ( t ) ) ∈ [ 0 , 1 ] × [ 0 , R ] × [ 0 , R ] , there exists K > 0 , such that f ( s , u ( s ) , D 0 + β u ( s ) ) ≤ K . Let

L 1 = λ ∫ 0 1 G ( 1 , s ) d s ; L 2 = λ ∫ 0 1 H ( 1 , s ) d s ,

we have

‖ Au ‖ 1 = max t ∈ [ 0 , 1 ] ∣ ( Au ) ( t ) ∣ = λ max t ∈ [ 0 , 1 ] ∫ 0 1 G ( t , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≤ K λ ∫ 0 1 G ( 1 , s ) d s = K L 1 ,

and

‖ Au ‖ 2 = max t ∈ [ 0 , 1 ] ∣ D 0 + β ( Au ) ( t ) ∣ = λ max t ∈ [ 0 , 1 ] ∫ 0 1 H ( t , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≤ K λ ∫ 0 1 H ( 1 , s ) d s = K L 2 .

Then

‖ Au ‖ = ‖ Au ‖ 1 + ‖ Au ‖ 2 ≤ K ( L 1 + L 2 ) .

Hence, A ( Ω ) is uniformly bounded.

A : P → P is equicontinuous.

Since G ( t , s ) and H ( t , s ) are continuous on [ 0 , 1 ] × [ 0 , 1 ] , we know G ( t , s ) and H ( t , s ) are uniformly continuous on [ 0 , 1 ] × [ 0 , 1 ] . Take t 1 , t 2 , s ∈ [ 0 , 1 ] , for any ε > 0 , there exists a constant δ > 0 , whenever ∣ t 2 − t 1 ∣ < δ , we have

∣ G ( t 2 , s ) − G ( t 1 , s ) ∣ < ε 2 K λ + 1

and

∣ H ( t 2 , s ) − H ( t 1 , s ) ∣ < ε 2 K λ + 1 .

Furthermore,

∣ ( Au ) ( t 2 ) − ( Au ) ( t 1 ) ∣ ≤ λ ∫ 0 1 ∣ G ( t 2 , s ) − G ( t 1 , s ) ∣ ∣ f ( s , u ( s ) , D 0 + β u ( s ) ) ∣ d s ≤ K λ ∫ 0 1 ∣ G ( t 2 , s ) − G ( t 1 , s ) ∣ d s < ε 2

and

∣ D 0 + β ( Au ) ( t 2 ) − D 0 + β ( Au ) ( t 1 ) ∣ ≤ λ ∫ 0 1 ∣ H ( t 2 , s ) − H ( t 1 , s ) ∣ ∣ f ( s , u ( s ) , D 0 + β u ( s ) ) ∣ d s ≤ K λ ∫ 0 1 ∣ H ( t 2 , s ) − H ( t 1 , s ) ∣ d s < ε 2 .

Thus,

‖ ( Au ) ( t 2 ) − ( Au ) ( t 1 ) ‖ < ε .

According to the Arzela-Ascoli theorem, we can show that A is relatively compact, which implies that the operator A is completely continuous.□

4 Existence of one positive solution

In this section, we give some sufficient conditions for the existence of at least one positive solution, and as proof, we give several simulation examples.

For convenience, we denote

k 1 = Γ ( α ) + Γ ( α − β ) Γ ( α ) Γ ( α − β ) , k 2 = Γ ( α + 1 ) + Γ ( α − β + 1 ) Γ ( α + 1 ) Γ ( α − β + 1 ) , f ∞ = lim ( u , v ) → + ∞ sup t ∈ [ 0 , 1 ] f ( t , u , v ) u + v , f 0 = lim ( u , v ) → 0 sup t ∈ [ 0 , 1 ] f ( t , u , v ) u + v , f ∞ = lim ( u , v ) → + ∞ inf t ∈ [ 0 , 1 ] f ( t , u , v ) u + v , f 0 = lim ( u , v ) → 0 inf t ∈ [ 0 , 1 ] f ( t , u , v ) u + v , M = λ − 1 g ( 0 ) k 1 ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) , N = 1 4 1 − α λ − 1 g ( 0 ) k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) .

Theorem 4.1

Assume f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , if there exists two constants 0 < r 2 < r 1 , and the following conditions hold:

f ( t , u , D 0 + β u ) ≤ r 1 M , for ( t , u , D 0 + β u ) ∈ [ 0 , 1 ] × [ 0 , r 1 ] × [ 0 , r 1 ] ;
f ( t , u , D 0 + β u ) ≥ r 2 N , for ( t , u , D 0 + β u ) ∈ 1 4 , 1 × [ 0 , r 2 ] × [ 0 , r 2 ] ;

then, the m -point fractional differential equation (1.1) has at least one positive solution.

Proof

Let

Ω r 1 = { u ∈ P : ‖ u ‖ < r 1 } ,

for any u ∈ P ∩ ∂ Ω r 1 , ‖ u ‖ = r 1 . By the condition (H1), we have

‖ Au ‖ = max t ∈ [ 0 , 1 ] ∣ ( Au ) ( t ) ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β ( Au ) ( t ) ∣ = λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s

≤ λ r 1 M ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 g ( 0 ) Γ ( α ) d s + 1 Γ ( α + 1 ) + ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 g ( 0 ) Γ ( α − β ) d s + 1 Γ ( α − β + 1 ) ≤ r 1 M k 1 ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) λ − 1 g ( 0 ) = r 1 ,

which implies that ‖ Au ‖ ≤ ‖ u ‖ for u ∈ P ∩ ∂ Ω r 1 .

Let

Ω r 2 = { u ∈ P : ‖ u ‖ < r 2 } ,

for any u ∈ P ∩ ∂ Ω r 2 , ‖ u ‖ = r 2 . By condition (H2), we have

‖ Au ‖ = max t ∈ [ 0 , 1 ] ∣ ( Au ) ( t ) ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β ( Au ) ( t ) ∣ ≥ λ 1 4 α − 1 ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ 1 4 α − β − 1 ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ 1 4 α − 1 r 2 N ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 g ( 0 ) Γ ( α ) d s + 1 Γ ( α + 1 ) + ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 g ( 0 ) Γ ( α − β ) d s + 1 Γ ( α − β + 1 ) ≥ r 2 N 1 4 α − 1 k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) λ − 1 g ( 0 ) = r 2 .

Then

‖ Au ‖ ≥ ‖ u ‖ .

According to Lemma 2.9, the m -point boundary value problem (1.1) has at least one solution u ∈ P ∩ ( Ω ¯ r 1 ⧹ Ω r 2 ) .□

Example 1

Consider the following boundary value problem:

(4.1) D 0 + 7 2 u ( t ) + λ f t , u ( t ) , D 0 + 3 2 u ( t ) = 0 , t ∈ [ 0 , 1 ] , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 2 ) ( 0 ) = 0 , D 0 + 3 2 u ( 1 ) = 1 4 D 0 + 3 2 u 1 5 + 1 2 D 0 + 3 2 u 1 4 ,

where α = 7 2 , β = 3 2 , m = 4 , ε 1 = 1 4 , η 1 = 1 5 , ε 2 = 1 2 , η 2 = 1 4 , λ = 1 , and

f t , u , u 3 2 = 500 − u − u 3 2 + sin t .

Then α − β = 2 > 1 , g ( 0 ) = 0.825 , and

k 1 = Γ 7 2 + Γ ( 2 ) Γ 7 2 Γ ( 2 ) ≈ 1.3008 , k 2 = Γ 9 2 + Γ ( 3 ) Γ 9 2 Γ ( 3 ) ≈ 0.58596 ,

M = λ − 1 g ( 0 ) k 1 ∫ 0 1 g ( s ) ( 1 − s ) d s + k 2 g ( 0 ) ≈ 0.5918 , N = 4 5 2 λ − 1 g ( 0 ) k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) d s + k 2 g ( 0 ) ≈ 23.2426 .

Let r 1 = 1,000 , r 2 = 1 , then 0 < r 2 < r 1 and

500 − u − u 3 2 + sin t = 498 > 23.2426 = r 2 N , t , u , u 3 2 ∈ 1 4 , 1 × [ 0 , 1 ] × [ 0 , 1 ] ; 500 − u − u 3 2 + sin t = 500 + sin 1 < 591.8 = r 1 M , t , u , u 3 2 ∈ [ 0 , 1 ] × [ 0 , 1,000 ] × [ 0 , 1,000 ] .

Thus, all the conditions of Theorem 4.1 are satisfied. By Theorem 4.1, the fractional differential equation (4.1) has at least one positive solution.

Now, we use the iterative method similar to that proposed by Wei et al. in [18] to simulate this process. Firstly, given θ ∈ C [ 0 , 1 ] , let

u ( t ) = λ ∫ 0 1 G ( t , s ) θ ( s ) d s , D 0 + 3 2 u ( t ) = λ ∫ 0 1 H ( t , s ) θ ( s ) d s .

Define the operator A : C [ 0 , 1 ] → C [ 0 , 1 ] , by

( A θ ) ( t ) = f t , λ ∫ 0 1 G ( t , s ) θ ( s ) d s , λ ∫ 0 1 H ( t , s ) θ ( s ) d s .

According to the continuity of G ( t , s ) and θ ( s ) , we can know A is the continuous operator. By the definition of Green function, it is easy to prove θ is a fixed point of operator A .

Then

u ( t ) = − 1 Γ 7 2 ∫ 0 t ( t − s ) 5 2 θ ( s ) d s + t 5 2 g ( 0 ) Γ 7 2 ∫ 0 1 ( 1 − s ) − 1 2 1 5 − s − 1 2 1 4 − s θ ( s ) d s

is a solution to problem (4.1).

Let

θ 0 ( t ) = f ( t , 0 , 0 ) = 500 + sin t , θ k + 1 ( t ) = f t , λ ∫ 0 1 G ( t , s ) θ k ( s ) d s , λ ∫ 0 1 H ( t , s ) θ k ( s ) d s .

From the proof of Wei et al. in [18], we know the operator A is contraction and this iterative method converges with the rate of geometric progression. Therefore, u ( t ) = λ ∫ 0 1 G ( t , s ) θ ∗ ( s ) d s is a solution of problem (4.1), where θ ∗ is a fixed point of operator A .

Now, we give the iterative process and approximate solution (Figures 1 and 2).

$Figure 1 The iterative process.$

Figure 1

The iterative process.

$Figure 2 The approximate solution.$

Figure 2

The approximate solution.

Theorem 4.2

Assume that there exist two constants Λ 1 , Λ 2 > 0 , if f ∞ < Λ 1 , f 0 > Λ 2 ,

Λ 1 < 4 1 − α k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) k 1 ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) Λ 2 ,

and λ satisfies

1 4 1 − α g ( 0 ) Λ 2 − 1 k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) ≤ λ ≤ g ( 0 ) Λ 1 − 1 k 1 ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) ,

the m -point fractional differential equation (1.1) has at least one positive solution.

Proof

According to f 0 > Λ 2 , there exists m 2 > 0 , such that for 0 < u < m 2 , there is

f ( t , u , D 0 + β u ) ≥ Λ 2 ( u + D 0 + β u ) ≥ Λ 2 1 4 α − 1 ‖ u ‖ .

Let

Ω m 2 = { u ∈ P : ‖ u ‖ < m 2 } ,

for any u ∈ P ∩ ∂ Ω m 2 , we have

‖ Au ‖ = max t ∈ [ 0 , 1 ] ∣ ( Au ) ( t ) ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β ( Au ) ( t ) ∣ = λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ ‖ u ‖ k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) Λ 2 − 1 1 4 1 − α g ( 0 ) ≥ ‖ u ‖ .

Then

‖ Au ‖ ≥ ‖ u ‖ .

Since f ∞ < Λ 1 , there exists m 1 > m 2 > 0 , such that

f ( t , u , D 0 + β u ) ≤ Λ 1 ( u + D 0 + β u ) ≤ Λ 1 ( ‖ u ‖ 1 + ‖ u ‖ 2 ) ≤ Λ 1 ‖ u ‖ ,

for ( t , u , D 0 + β u ) ∈ [ 0 , 1 ] × [ m 1 , + ∞ ] × [ m 1 , + ∞ ] .

Let

Ω m 1 = { u ∈ P : ‖ u ‖ < m 1 } ,

for any u ∈ P ∩ ∂ Ω m 1 , we have

‖ Au ‖ = max t ∈ [ 0 , 1 ] ∣ ( Au ) ( t ) ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β ( Au ) ( t ) ∣ = λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≤ λ ‖ u ‖ k 1 ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) Λ 1 − 1 g ( 0 ) ≤ ‖ u ‖ .

Then

‖ Au ‖ ≤ ‖ u ‖ .

According to Lemma 2.9, the m -point fractional differential equation (1.1) has one positive solution u ∈ P ∩ ( Ω ¯ m 1 ⧹ Ω m 2 ) .□

Theorem 4.3

Assume f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , f 0 = + ∞ , f ∞ = 0 , the m -point boundary value problem (1.1) has at least one positive solution.

Proof

Since f 0 = + ∞ , there exists h 1 > 0 such that

f ( t , u 1 , u 2 ) ≥ σ 1 ( u 1 + u 2 ) ≥ σ 1 1 4 α − 1 ‖ u ‖ ,

for ( t , u 1 , u 2 ) ∈ 1 4 , 1 × [ 0 , h 1 ] × [ 0 , h 1 ] , where

σ 1 ≥ 4 α − 1 λ ∫ 0 1 G ( 1 , s ) d s + λ ∫ 0 1 H ( 1 , s ) d s − 1 .

We have

‖ Au ‖ = max t ∈ [ 0 , 1 ] ∣ ( Au ) ( t ) ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β ( Au ) ( t ) ∣ = λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ σ 1 1 4 α − 1 ‖ u ‖ ∫ 0 1 G ( 1 , s ) d s + ∫ 0 1 H ( 1 , s ) d s ≥ ‖ u ‖ .

Then

‖ Au ‖ ≥ ‖ u ‖ .

On the other hand, since f ∞ = 0 , there exists h 2 > h 1 > 0 such that

f ( t , u 1 , u 2 ) ≤ σ 2 ( u 1 + u 2 ) ≤ σ 2 ‖ u ‖ ,

for ( t , u 1 , u 2 ) ∈ [ 0 , 1 ] × [ h 2 , + ∞ ) × [ h 2 , + ∞ ) , where

σ 2 ≤ λ ∫ 0 1 G ( 1 , s ) d s + λ ∫ 0 1 H ( 1 , s ) d s − 1 .

We have

‖ Au ‖ = max t ∈ [ 0 , 1 ] ∣ ( Au ) ( t ) ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β ( Au ) ( t ) ∣ = λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≤ λ σ 2 ‖ u ‖ ∫ 0 1 G ( 1 , s ) d s + ∫ 0 1 H ( 1 , s ) d s ≤ ‖ u ‖ .

Then

‖ Au ‖ ≤ ‖ u ‖ .

Consequently, the m -point fractional differential equation (1.1) has at least one positive solution u ∈ P ∩ ( Ω ¯ h 2 ⧹ Ω h 1 ) .□

Theorem 4.4

For f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , assume f 0 = 0 , f ∞ = + ∞ , the m -point boundary value problem (1.1) has at least one positive solution u ∈ P .

Proof

Under the above assumptions, by using the similar way in the proof of Theorem 4.3, it is easy to prove that Lemma 2.9 is satisfied. So we omit the proof.□

Example 2

Consider the following boundary value problem:

(4.2) D 0 + 10 3 u ( t ) + λ f t , u ( t ) , D 0 + 4 3 u ( t ) = 0 , t ∈ [ 0 , 1 ] , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 2 ) ( 0 ) = 0 , D 0 + 4 3 u ( 1 ) = 1 3 D 0 + 4 3 u 1 4 + 3 4 D 0 + 4 3 u 1 2 ,

where α = 10 3 , β = 4 3 , m = 4 , ε 1 = 1 3 , η 1 = 1 4 , ε 2 = 3 4 , η 2 = 1 2 .

Case 1:

f t , u , u 4 3 = u + u 4 3 + t , t , u , u 4 3 ∈ [ 0 , 1 ] × [ 0 , 1 ] × [ 0 , 1 ] , 1 1,000 ( u + u 4 3 ) + 999 500 + t , t , u , u 4 3 ∈ [ 0 , 1 ] × ( 1 , + ∞ ) × ( 1 , + ∞ ) .

We can obtain α − β = 2 > 1 , g ( 0 ) = 0.54167 , k 1 ≈ 1.3601 , k 2 ≈ 0.60802 , and

g ( 0 ) k 1 ∫ 0 1 g ( s ) ( 1 − s ) d s + k 2 g ( 0 ) ≈ 0.48246 , 4 7 3 g ( 0 ) k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) d s + k 2 g ( 0 ) ≈ 14.714 .

Let Λ 1 = 1 999 , Λ 2 = 1 2 , we have f ∞ = 1 1,000 < 1 999 , f 0 = 1 > 1 2 ,

1 999 = Λ 1 < 4 − 7 3 k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) d s + k 2 g ( 0 ) k 1 ∫ 0 1 g ( s ) ( 1 − s ) d s + k 2 g ( 0 ) Λ 2 = 0.03279 Λ 2 = 0.01639 ,

and λ 1 , λ 2 satisfy 29.428 ≤ λ ≤ 482.46 .

Thus, all the conditions of Theorem 4.2 are satisfied. By Theorem 4.2, the fractional differential equation (4.2) has at least one positive solution.

For

f t , u , u 4 3 = u + u 4 3 + t ,

select the initial value as follows:

θ 0 ( t ) = f ( t , 0 , 0 ) = t .

The iteration formula is

θ k + 1 ( t ) = f t , λ ∫ 0 1 G ( t , s ) θ k ( s ) d s , λ ∫ 0 1 H ( t , s ) θ k ( s ) d s .

By using the iterative method of Example 1, the iterative process and approximate solution are given in Figures 3 and 4.

Case 2:

f t , u , u 4 3 = 1 + sin t u 4 3 + 2 u , t , u , u 4 3 ∈ [ 0 , 1 ] × [ 0 , + ∞ ) × [ 0 , + ∞ ) ,

it is easy to obtain that f 0 = + ∞ , f ∞ = 0 . According to Theorem 4.3, the m -point boundary value problem (4.2) has at least one positive solution.

Let

θ 0 ( t ) = f ( t , 0 , 0 ) = 1 + sin t , θ k + 1 ( t ) = f t , λ ∫ 0 1 G ( t , s ) θ k ( s ) d s , λ ∫ 0 1 H ( t , s ) θ k ( s ) d s .

The iterative process and iterative solution are as follows (Figures 5 and 6).

$Figure 3 The iterative process.$

Figure 3

The iterative process.

$Figure 4 The approximate solution.$

Figure 4

The approximate solution.

$Figure 5 The iterative process.$

Figure 5

The iterative process.

$Figure 6 The approximate solution.$

Figure 6

The approximate solution.

5 Existence of multiple positive solutions

In this section, by using multiple fixed point theorems and mathematical induction, we give some sufficient conditions for the existence of at least 2, 3, and 2 n − 1 positive solutions.

Theorem 5.1

Assume f 0 = f ∞ = + ∞ and exists v 2 > 0 such that

f ( t , u , D 0 + β u ) ≤ v 2 M , ( t , u , D 0 + β u ) ∈ [ 0 , 1 ] × [ 0 , v 2 ] × [ 0 , v 2 ] .

Then, the m -point boundary value problem of fractional differential equation (1.1) has at least two positive solutions.

Proof

According to f ( t , u , D 0 + β u ) ≤ v 2 M , we set

Ω v 2 = { u ∈ P : ‖ u ‖ < v 2 } ,

for any u ∈ P ∩ ∂ Ω v 2 , there is ‖ u ‖ = v 2 , we can obtain

‖ Au ‖ ≤ λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≤ v 2 M k 1 ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) λ − 1 g ( 0 ) = v 2 .

Then

‖ Au ‖ ≤ ‖ u ‖ .

Since f 0 = + ∞ , there exists 0 < v 1 < v 2 , such that

f ( t , u , D 0 + β u ) ≥ σ 1 ( u + D 0 + β u ) ≥ σ 1 1 4 α − 1 ‖ u ‖ ,

for ( t , u 1 , u 2 ) ∈ 1 4 , 1 × [ 0 , v 1 ] × [ 0 , v 1 ] .

Let

Ω v 1 = { u ∈ P : ‖ u ‖ < v 1 } .

By using the similar way to the proof of Theorem 4.3, we have ‖ Au ‖ ≥ ‖ u ‖ for u ∈ ∂ Ω v 1 .

For f ∞ = + ∞ , there exists a constant v 3 > v 2 , such that

f ( t , u , D 0 + β u ) ≥ σ 1 ( u + D 0 + β u ) ≥ σ 1 1 4 α − 1 ‖ u ‖ ,

for ( t , u 1 , u 2 ) ∈ 1 4 , 1 × [ v 3 , + ∞ ) × [ v 3 , + ∞ ) .

Let

Ω v 3 = { u ∈ P : ‖ u ‖ < v 3 } .

By using the similar way to the proof of Theorem 4.3, we have ‖ Au ‖ ≥ ‖ u ‖ for u ∈ ∂ Ω v 3 .

Then, all the conditions of Lemma 2.9 are satisfied, it implies that m -point boundary value problem of fractional differential equation (1.1) has at least two positive solutions u 1 , u 2 and u 1 ∈ P ∩ ( Ω ¯ v 2 ⧹ Ω v 1 ) , u 2 ∈ P ∩ ( Ω ¯ v 3 ⧹ Ω v 2 ) .□

In a similar way, we have the following result.

Theorem 5.2

Assume f 0 = f ∞ = 0 and exists φ > 0 such that

f ( t , u 1 , u 2 ) ≥ φ N , for ( t , u 1 , u 2 ) ∈ 1 4 , 1 × [ 0 , φ ] × [ 0 , φ ] .

Then, the m -point boundary value problem of fractional differential equation (1.1) has at least two positive solutions.

Example 3

Consider the following boundary value problem:

(5.1) D 0 + 11 3 u ( t ) + λ f t , u ( t ) , D 0 + 4 3 u ( t ) = 0 , t ∈ [ 0 , 1 ] , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 2 ) ( 0 ) = 0 , D 0 + 4 3 u ( 1 ) = 1 10 D 0 + 4 3 u ( 5 7 ) ,

where α = 11 3 , β = 4 3 , m = 3 , ε 1 = 1 10 , η 1 = 5 7 ,

f t , u , u 4 3 = 8 + ( u + u 4 3 ) 2 + t 2 M , t , u , u 4 3 ∈ [ 0 , 1 ] × [ 0 , 1 ] × [ 0 , 1 ] , 12 + t 2 M , t , u , u 4 3 ∈ [ 0 , 1 ] × ( 1 , + ∞ ) × ( 1 , + ∞ ) .

We have α − β = 7 3 > 1 , f 0 = + ∞ , f ∞ = + ∞ , M = 0.47701 . Let υ 2 = 2 , it is easy to obtain f t , u , u 4 3 ≤ 25 2 M for t , u , u 4 3 ∈ 1 4 , 1 × [ 0 , 2 ] × [ 0 , 2 ] .

Thus, all the conditions of Theorem 5.1 are satisfied. By Theorem 5.1, the fractional differential equation (5.1) has at least two positive solutions.

By using the iterative method of Example 1, we obtain the following approximate solutions (Figures 7 and 8).

$Figure 7 The approximate solution u 1 {u}_{1} .$

Figure 7

The approximate solution u 1 .

$Figure 8 The approximate solution u 2 {u}_{2} .$

Figure 8

The approximate solution u 2 .

Theorem 5.3

Assume f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , there exist three constants 0 < a 1 < b 1 < c 1 , and satisfies the following conditions:

f ( t , u , D 0 + β u ) > a 1 M , for ( t , u , D 0 + β u ) ∈ [ 0 , 1 ] × [ 0 , a 1 ] × [ 0 , a 1 ] ;
f ( t , u , D 0 + β u ) < b 1 M , for ( t , u , D 0 + β u ) ∈ [ 0 , 1 ] × [ 0 , b 1 ] × [ 0 , b 1 ] ;
f ( t , u , D 0 + β u ) > c 1 N , for ( t , u , D 0 + β u ) ∈ 1 4 , 1 × [ 0 , H c 1 ] × [ 0 , H c 1 ] .

Then, the m -point fractional differential equation (1.1) has at least two positive solutions u 1 , u 2 and

a 1 < α 1 ( u 1 ) , θ 1 ( u 1 ) < b 1 < θ 1 ( u 2 ) , γ 1 ( u 2 ) < c 1 .

Proof

Let

γ 1 ( u ) = min t ∈ 1 4 , 1 ∣ u ( t ) ∣ + min t ∈ 1 4 , 1 ∣ D 0 + β u ( t ) ∣ , θ 1 ( u ) = α 1 ( u ) = max t ∈ [ 0 , 1 ] ∣ u ( t ) ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β u ( t ) ∣ .

Obviously, α 1 ( u ) , θ 1 ( u ) , and γ 1 ( u ) are increasing, nonnegative continuous functionals, and γ 1 ( u ) ≤ θ 1 ( u ) = α 1 ( u ) , θ 1 ( 0 ) = 0 .

Since u ( t ) = ∫ 0 1 G ( t , s ) y ( s ) d s , we have

γ 1 ( u ) = min t ∈ 1 4 , 1 ∣ u ( t ) ∣ + min t ∈ 1 4 , 1 ∣ D 0 + β u ( t ) ∣ = λ ∫ 0 1 G 1 4 , s f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H 1 4 , s f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ 1 4 α − 1 ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ 1 4 α − 1 ( ‖ u ‖ 1 + ‖ u ‖ 2 ) = 1 4 α − 1 ‖ u ‖ .

Hence,

(5.2) ‖ u ‖ ≤ 4 α − 1 γ 1 ( u ) = H γ 1 ( u ) .

Then, the condition (i) of Lemma 2.10 is satisfied.

According to (H5), by using the similar way of Lemma 3.1, we can obtain the operator A : P ¯ ( γ 1 , c 1 ) → P . Then condition (iii) of Lemma 2.10 is satisfied.

For any ψ ∈ [ 0 , 1 ] , u ∈ ∂ P ( θ 1 , b 1 ) , we have

θ 1 ( ψ u ) = max t ∈ [ 0 , 1 ] ( ψ u ( t ) ) + max t ∈ [ 0 , 1 ] ( ψ D 0 + β u ( t ) ) = ψ max t ∈ [ 0 , 1 ] u ( t ) + ψ max t ∈ [ 0 , 1 ] D 0 + β u ( t ) = ψ θ 1 ( u ) .

This implies condition (ii) of Lemma 2.10 is satisfied.

By condition (H3), for u ∈ ∂ P ( α 1 , a 1 ) , we can obtain α 1 ( u ) = a 1 , 0 ≤ u ( t ) ≤ ‖ u ‖ ≤ a 1 , 0 ≤ D 0 + β u ( t ) ≤ a 1 , and

α 1 ( Au ) = max t ∈ [ 0 , 1 ] ∣ Au ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β Au ∣ = λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s > λ a 1 M ∫ 0 1 G ( 1 , s ) d s + ∫ 0 1 H ( 1 , s ) d s = a 1 ,

it implies that α 1 ( Au ) > a 1 , for u ∈ ∂ P ( α 1 , a 1 ) .

Let u = a 1 2 , then

α 1 ( u ) = α 1 a 1 2 = a 1 2 < a 1 .

Hence, P ( α 1 , a 1 ) ≠ ∅ .

By (H4), for u ∈ ∂ P ( θ 1 , b 1 ) , we can obtain θ 1 ( u ) = b 1 , then 0 ≤ u ( 1 ) ≤ ‖ u ‖ ≤ b 1 , 0 ≤ D 0 + β u ( t ) ≤ b 1 , and

θ 1 ( Au ) = max t ∈ [ 0 , 1 ] ∣ Au ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β Au ∣ = λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s < λ b 1 M ∫ 0 1 G ( 1 , s ) d s + λ ∫ 0 1 H ( 1 , s ) d s = b 1 .

Then, θ 1 ( Au ) < b 1 , for u ∈ ∂ K ( θ 1 , b 1 ) .

According to (H5), for u ∈ ∂ P ( γ 1 , c 1 ) , we can obtain γ 1 ( u ) = c 1 , then 0 ≤ u ( t ) ≤ ‖ u ‖ ≤ H γ 1 ( u ) ≤ H c 1 , 0 ≤ D 0 + β u ( t ) ≤ H c 1 , and

γ 1 ( Au ) = min t ∈ 1 4 , 1 ∣ Au ∣ + min t ∈ 1 4 , 1 ∣ D 0 + β Au ∣ = λ ∫ 0 1 G ( 1 4 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + ∫ 0 1 H 1 4 , s f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ c 1 N 1 4 α − 1 ∫ 0 1 G ( 1 , s ) d s + ∫ 0 1 H ( 1 , s ) d s > c 1 N k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) 1 4 1 − α λ − 1 g ( 0 ) = c 1 .

Thus, γ 1 ( Au ) > c 1 , for u ∈ ∂ P ( γ 1 , c 1 ) .

To sum up, all conditions of Lemma 2.10 are satisfied. So, the m -point fractional differential equation (1.1) has at least two positive solutions u 1 , u 2 , and

a 1 < α 1 ( u 1 ) , θ 1 ( u 1 ) < b 1 < θ 1 ( u 2 ) , γ 1 ( u 2 ) < c 1 . □

With a small change to Lemma 2.10, Avery et al. obtained Lemma 2.11. Then, we have the following result.

Theorem 5.4

Assume f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , there exist three numbers a 2 , b 2 , c 2 with 0 < a 2 < b 2 < c 2 , and b 2 N < c 2 M . If f satisfies the following conditions:

f ( t , u , D 0 + β u ) < a 2 M , for ( t , u , D 0 + β u ) ∈ [ 0 , 1 ] × [ 0 , a 2 ] × [ 0 , a 2 ] ;
f ( t , u , D 0 + β u ) > b 2 N , for ( t , u , D 0 + β u ) ∈ 1 4 , 1 × [ 0 , H b 2 ] × [ 0 , H b 2 ] ;
f ( t , u , D 0 + β u ) < c 2 M , for ( t , u , D 0 + β u ) ∈ 1 4 , 1 × [ 0 , H c 2 ] × [ 0 , H c 2 ] ;

then the m -point boundary value problem of fractional differential equation (1.1) has at least three positive solutions.

Proof

For u ∈ P ¯ , we define the following functions:

γ 2 ( u ) = θ 2 ( u ) = min t ∈ 1 4 , 1 ∣ u ( t ) ∣ + min t ∈ 1 4 , 1 ∣ D 0 + β u ( t ) ∣ , α 2 ( u ) = max t ∈ [ 0 , 1 ] ∣ u ( t ) ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β u ( t ) ∣ ,

for u ∈ P . α 2 ( u ) , θ 2 ( u ) , and γ 2 ( u ) be increasing, nonnegative continuous functionals.

According to (5.2), we have

‖ u ‖ ≤ 4 α − 1 γ 2 ( u ) = H γ 2 ( u ) = H θ 2 ( u ) .

By using the similar way to the proof of Lemma 3.1, we can obtain

A : P ¯ ( γ 2 , c 2 ) → P .

Now, we show all the conditions of Lemma 2.11 are satisfied.

For u ∈ ∂ P ( γ 2 , c 2 ) , we have

γ 2 ( u ) = c 2 , 0 ≤ u ( t ) ≤ ‖ u ‖ ≤ H γ 2 ( u ) ≤ H c 2 , 0 ≤ D 0 + β u ( t ) ≤ ‖ u ‖ ≤ H c 2 .

By (H8), we obtain

γ 2 ( Au ) = min t ∈ 1 4 , 1 ∣ Au ∣ + min t ∈ 1 4 , 1 ∣ D 0 + β Au ∣ = λ ∫ 0 1 G 1 4 , s f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H 1 4 , s f ( s , u ( s ) , D 0 + β u ( s ) ) d s < λ c 2 M ∫ 0 1 G ( 1 , s ) d s + λ ∫ 0 1 H ( 1 , s ) d s < λ c 2 M k 1 ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) g ( 0 ) = c 2 .

Let u = a 1 3 , then

α 2 ( u ) = α 2 a 1 3 = a 1 3 < a 2 .

Hence, P ( α 2 , a 2 ) ≠ ∅ .

For u ∈ ∂ P ( α 2 , a 2 ) , we have

α 2 ( u ) = a 2 , 0 ≤ u ( t ) ≤ ‖ u ‖ ≤ a 2 , 0 ≤ D 0 + β u ( t ) ≤ ‖ u ‖ ≤ a 2 .

It follows from the condition of (H6), we can obtain

α 2 ( Au ) = max t ∈ [ 0 , 1 ] ∣ Au ∣ + max t ∈ [ 0 , 1 ] ∣ D 0 + β Au ∣ = λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s < λ a 2 M ∫ 0 1 G ( 1 , s ) d s + λ ∫ 0 1 H ( 1 , s ) d s < λ a 2 M k 1 ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) g ( 0 ) = a 2 .

For u ∈ ∂ P ( θ 2 , b 2 ) , we have

θ 2 ( u ) = b 2 , 0 ≤ u ( t ) ≤ ‖ u ‖ ≤ H θ 2 ( u ) ≤ H b 2 , 0 ≤ D 0 + β u ( t ) ≤ ‖ u ‖ ≤ H b 2 .

According to (H7), we show that

θ 2 ( Au ) = min t ∈ 1 4 , 1 ∣ Au ∣ + min t ∈ 1 4 , 1 ∣ D 0 + β Au ∣ = λ ∫ 0 1 G 1 4 , s f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H 1 4 , s f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ b 1 ′ N 1 4 α − 1 ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + λ ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s

> b 2 N k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) 1 4 1 − α λ − 1 g ( 0 ) = b 2 .

Then, all conditions of Lemma 2.11 are satisfied. The boundary value problems (1.1) has at least three positive solutions u 1 , u 2 , and u 3 belonging to P ( γ 2 , c 2 ) .□

Theorem 5.5

Assume f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , there exists 0 < a 3 < b 3 < d < c 3 , and 1 4 1 − α b 3 λ ∫ 0 1 G ( 1 , s ) d s < c 3 M . If f satisfies the following conditions:

f ( t , u , D 0 + β u ) < a 3 M , for ( t , u , D 0 + β u ) ∈ [ 0 , 1 ] × [ 0 , a 3 ] × [ 0 , a 3 ] ;
f ( t , u , D 0 + β u ) > 1 4 1 − α b 3 λ ∫ 0 1 G ( 1 , s ) d s , for ( t , u , D 0 + β u ) ∈ 1 4 , 3 4 × [ b 3 , c 3 ] × [ 0 , c 3 ] ;
f ( t , u , D 0 + β u ) ≤ c 3 M , for ( t , u , D 0 + β u ) ∈ [ 0 , 1 ] × [ 0 , c 3 ] × [ 0 , c 3 ] ;

then, the m -point boundary value problem of fractional differential equation (1.1) has at least three positive solutions u 1 , u 2 , and u 3 such that

‖ u 1 ‖ < a , b < θ ( u 2 ) and u 3 > a with θ ( u 3 ) < b .

Proof

Let

P c 3 = { x ∈ P : ‖ u ‖ ≤ c 3 } , P ( θ , b 3 , d ) = { u ∈ P : b 3 ≤ θ ( u ) , ‖ u ‖ ≤ d } .

We define

θ ( u ) = min t ∈ 1 4 , 3 4 u ( t ) .

Then θ ( u ) is a nonnegative continuous concave function.

For u ∈ P ¯ c 3 , we have ‖ u ‖ ≤ c 3 , thus 0 ≤ u ( t ) ≤ ‖ u ‖ ≤ c 3 and 0 ≤ D 0 + β u ( t ) ≤ c 3 .

According to condition (H11), we have

‖ Au ‖ = λ ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s + ∫ 0 1 H ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≤ c 3 M k 1 ∫ 0 1 g ( s ) ( 1 − s ) α − β − 1 d s + k 2 g ( 0 ) λ − 1 g ( 0 ) = c 3 ,

which implies A : P ¯ c 3 → P ¯ c 3 .

Similarly, for u ∈ P ¯ a 3 , we have ‖ u ‖ ≤ a 3 , hence 0 ≤ u ( t ) ≤ ‖ u ‖ ≤ a 3 and 0 ≤ D 0 + β u ( t ) ≤ a 3 . We can obtain that ‖ Au ‖ < a 3 .

Thus, condition (ii) of Lemma 2.12 is satisfied.

Let u ( t ) = b 3 + c 3 2 ∈ P ( θ , b 3 , c 3 ) , we have

θ ( u ) = θ b 3 + c 3 2 > b 3 .

Then, { u ∈ P ( θ , b 3 , c 3 ) : θ ( u ) > b 3 } ≠ ∅ .

On the other hand, for u ∈ P ( θ , b 3 , c 3 ) , we can obtain

b 3 ≤ θ ( u ) ≤ u ( t ) ≤ ‖ u ‖ ≤ c 3 , 0 ≤ D 0 + β u ( t ) ≤ c 3 .

By (H10), we obtain

θ ( Au ) = min t ∈ [ 1 4 , 3 4 ] ∣ Au ∣ 1 = λ ∫ 0 1 G 1 4 , s f ( s , u ( s ) , D 0 + β u ( s ) ) d s ≥ λ 1 4 α − 1 ∫ 0 1 G ( 1 , s ) f ( s , u ( s ) , D 0 + β u ( s ) ) d s > λ 1 4 α − 1 1 4 1 − α b 3 λ ∫ 0 1 G ( 1 , s ) d s ∫ 0 1 G ( 1 , s ) d s = b 3 .

Then u ∈ P ( θ , b 3 , c 3 ) and ‖ Au ‖ > d , θ ( Au ) > b 3 , condition (iii) of Lemma 2.12 is satisfied. By using the same way, we can obtain θ ( Au ) > b 3 for u ∈ P ( θ , b 3 , d ) .

Consequently, all conditions of Lemma 2.12 are satisfied. The boundary value problem (1.1) has at least three positive solutions.□

Example 4

Consider the following boundary value problem:

(5.3) D 0 + 9 2 u ( t ) + λ f t , u ( t ) , D 0 + 3 2 u ( t ) = 0 , t ∈ [ 0 , 1 ] , u ( 0 ) = u ′ ( 0 ) = ⋯ = u ( n − 2 ) ( 0 ) = 0 , D 0 + 3 2 u ( 1 ) = 1 4 D 0 + 3 2 u 1 5 + 1 2 D 0 + 3 2 u 1 4 ,

where α = 9 2 , β = 3 2 , m = 4 , ε 1 = 1 4 , η 1 = 1 5 , ε 2 = 1 2 , η 2 = 1 4 , λ = 1 , and

f t , u , u 3 2 = 2 u , u ∈ [ 0 , 1 ] , 2 ( 700 u − 699 ) , u ∈ ( 1 , 2 ] , 7,002 , u ∈ ( 2 , + ∞ ) .

Then α − β = 3 > 1 , g ( 0 ) = 0.9586 , and

k 1 = Γ 9 2 + Γ ( 3 ) Γ 9 2 Γ ( 3 ) ≈ 0.58596 , k 2 = Γ 11 2 + Γ ( 4 ) Γ 11 2 Γ ( 4 ) ≈ 0.18577 , M = λ − 1 g ( 0 ) k 1 ∫ 0 1 g ( s ) ( 1 − s ) 2 d s + k 2 g ( 0 ) ≈ 3.1728 , N = 4 3.5 λ − 1 g ( 0 ) k 1 ∫ 1 4 1 g ( s ) ( 1 − s ) 2 d s + k 2 g ( 0 ) ≈ 638.1407 ,

1 4 1 − α b 3 λ ∫ 0 1 G ( 1 , s ) d s ≈ 3318.6414 .

Let a 3 = 1 , b 3 = 2 , c 3 = 2,500 , we see that 1 4 1 − α b 3 λ ∫ 0 1 G ( 1 , s ) d s < c 3 M and

f t , u , u 3 2 < a 3 M = 6.3456 , ( t , u 1 , u 2 ) ∈ [ 0 , 1 ] × [ 0 , 1 ] × [ 0 , 1 ] , f t , u , u 3 2 > 1 4 1 − α b 3 λ ∫ 0 1 G ( 1 , s ) d s = 6637.2828 , ( t , u 1 , u 2 ) ∈ 1 4 , 3 4 × [ 2 , 2,500 ] × [ 0 , 2,500 ] , f t , u , u 3 2 < c 3 M = 7,932 , ( t , u 1 , u 2 ) ∈ [ 0 , 1 ] × [ 0 , 2,500 ] × [ 0 , 2,500 ] .

Thus, all the conditions of Theorem 5.5 are satisfied. By Theorem 5.5, we can obtain that fractional differential equation (5.3) has at least three fixed points u 1 , u 2 , u 3 such that

‖ u 1 ‖ < a , b < q ( u 3 ) and u 3 > a with q ( u 3 ) < b .

Theorem 5.6

Assume f ∈ C ( [ 0 , 1 ] × [ 0 , + ∞ ) × [ 0 , + ∞ ) , [ 0 , + ∞ ) ) , there exist constant numbers a i ′ , b i ′ , and c i ′ such that 0 < a 1 ′ < b 1 ′ < c 1 ′ < a 2 ′ < b 2 ′ < c 2 ′ < ⋯ < a m ′ with

0 < b 1 ′ N < c 1 ′ M , … , b m − 1 ′ N < c m − 1 ′ M , n ∈ N ,

where i = 1 , … , m . If f satisfies the following conditions:

f ( t , u , D 0 + β u ) < a i ′ M , for ( t , u , D 0 + β u ) ∈ [ 0 , 1 ] × [ 0 , a i ′ ] × [ 0 , a i ′ ] ;
f ( t , u , D 0 + β u ) > b i ′ N , for ( t , u , D 0 + β u ) ∈ 1 4 , 1 × [ 0 , H b i ′ ] × [ 0 , H b i ′ ] ;
f ( t , u , D 0 + β u ) < c i ′ M , for ( t , u , D 0 + β u ) ∈ 1 4 , 1 × [ 0 , H c i ′ ] × [ 0 , H c i ′ ] ;

then, the m -point boundary value problem (1.1) has at least 2 n − 1 positive solutions.

Proof

We use mathematical induction to prove the result.

If m = 1 , from condition (i), we have A : P ¯ a 1 ′ → P a 1 ′ . According to the Schauder fixed point theorem, the boundary value problem (1.1) has at least one positive solution ‖ u ‖ ∈ P ¯ a 1 ′ .

For m = 2 , let a ′ = a 1 ′ , b ′ = b 1 ′ , c ′ = c 1 ′ . From Theorem 5.4, we obtain the boundary value problems (1.1) has at least three positive solutions u 1 , u 2 , and u 3 such that

0 < ‖ u 1 ‖ < a 1 ′ < ‖ u 2 ‖ , ‖ u 2 ‖ < b 1 ′ < ‖ u 3 ‖ , ‖ u 3 ‖ < c 1 ′ < a 2 ′ .

Assume that it is also true when m = n . Then, the boundary value problems (1.1) has at least 2 n − 1 positive solutions in P ¯ a n ′ .

We need to prove the result holds for m = n + 1 . By Theorem 5.4, the boundary value problem (1.1) has at least three positive solutions u , v , and w such that

0 < ‖ u ‖ < a n ′ < ‖ u ‖ , ‖ v ‖ < b n ′ < ‖ w ‖ , ‖ w ‖ < c n ′ < a n + 1 ′ .

Thus, the m -point boundary value problem (1.1) has at least 2 n + 1 positive solutions in P ¯ a n + 1 ′ , which implies that it is true for m = n + 1 .

Then, the m -point boundary value problem (1.1) has at least 2 n − 1 positive solutions.□

6 Conclusion

In this paper, we study the existence of positive solution for a class of m -point fractional differential equations whose nonlinear terms involve derivatives. By using the properties of the Green function and fixed point theorems, the existence of at least one, two, three, and 2 n − 1 positive solutions are established. As verification, we use the iterative method in [18] to simulate the examples, and give the shape of the approximate solution. In the following work, we believe that the idea of this paper can be extended to the coupled system, which is what we will study in the next step.

Acknowledgements

The authors express their sincere gratitude to the editors and referees for their helpful comments and suggestions.

Funding information: This work was partially supported by the Natural Science Foundation of China (12126427 and 12161079) and the Foundation of XZIT (XKY2020102).
Author contributions: All authors contributed equally and significantly in writing this article. All authors read and approved the final manuscript.
Conflict of interest: Authors state no conflict of interest.
Data availability statement：The datasets generated during the current study are available from the corresponding author on reasonable request.

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Received: 2021-02-04

Revised: 2021-12-07

Accepted: 2021-12-08

Published Online: 2021-12-31

This work is licensed under the Creative Commons Attribution 4.0 International License.

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https://doi.org/10.1515/math-2021-0131

Keywords for this article

fractional differential equations; existence; positive solution; simulation

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